Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int N;
  cin >> N;
  vector<int> x0, y0;
  vector<vector<int> > X(N, vector<int>(30));
  vector<vector<int> > Y = X;
  for (int i = 0; i < N; i++) {
    int x;
    cin >> x;
    x0.push_back(x);
    for (int j = 0; j < 30; j++) X[i][j] = (x >> j) & 1;
  }
  for (int i = 0; i < N; i++) {
    int x;
    cin >> x;
    y0.push_back(x);
    for (int j = 0; j < 30; j++) Y[i][j] = (x >> j) & 1;
  }
  vector<pair<int, int> > ans;
  vector<int> B;
  vector<bool> isB(N, false);
  for (int i = 0; i < 30; i++) {
    int x = -1;
    for (int j = 0; j < N; j++)
      if (X[j][i] == 1 && !isB[j]) x = j;
    if (x == -1) continue;
    B.push_back(x);
    isB[x] = true;
    for (int k = 0; k < N; k++)
      if (k != x && X[k][i] == 1) {
        ans.push_back(make_pair(k, x));
        for (int l = 0; l < 30; l++) X[k][l] ^= X[x][l];
      }
  }
  for (int i = 0; i < N; i++) {
    if (isB[i]) continue;
    for (int j = 0; j < 30; j++)
      if (Y[i][j] ^ X[i][j]) {
        bool ok = false;
        for (auto it = B.begin(); it != B.end(); it++)
          if (X[*it][j] == 1) {
            ans.push_back(make_pair(i, *it));
            for (int k = 0; k < 30; k++) X[i][k] ^= X[*it][k];
            ok = true;
          }
        if (!ok) {
          cout << "-1\n";
          return 0;
        }
      }
  }
  vector<pair<int, int> > ans2;
  int m = B.size();
  vector<vector<int> > M(m, vector<int>(m, 0));
  for (auto it = B.begin(); it != B.end(); it++) {
    for (int i = 0; i < m; i++)
      for (int j = 0; j < 30; j++)
        if (X[B[i]][j]) {
          if (!Y[*it][j]) break;
          M[it - B.begin()][i] = 1;
          for (int k = 0; k < 30; k++) Y[*it][k] ^= X[B[i]][k];
          break;
        }
    for (int i = 0; i < 30; i++)
      if (Y[*it][i]) {
        cout << "-1\n";
        return 0;
      }
  }
  vector<bool> piv(m, false);
  for (int i = 0; i < m; i++) {
    int x = -1;
    for (int j = 0; j < m; j++)
      if (M[j][i] && !piv[j]) x = j;
    if (x == -1) continue;
    piv[x] = true;
    for (int j = 0; j < m; j++)
      if (j != x && M[j][i]) {
        ans2.push_back(make_pair(B[j], B[x]));
        for (int k = 0; k < m; k++) M[j][k] ^= M[x][k];
      }
  }
  for (int i = 0; i < m; i++) {
    int x = -1;
    for (int j = i; j < m; j++)
      if (M[j][i]) x = j;
    if (x == -1 || x == i) continue;
    ans2.push_back(make_pair(B[i], B[x]));
    ans2.push_back(make_pair(B[x], B[i]));
    ans2.push_back(make_pair(B[i], B[x]));
    swap(M[i], M[x]);
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < i; j++)
      if (M[j][i]) ans.push_back(make_pair(B[j], B[i]));
    if (!M[i][i]) ans.push_back(make_pair(B[i], B[i]));
  }
  reverse(ans2.begin(), ans2.end());
  for (auto it = ans2.begin(); it != ans2.end(); it++) ans.push_back(*it);
  cout << ans.size() << "\n";
  for (auto it = ans.begin(); it != ans.end(); it++) {
    cout << it->first + 1 << " " << it->second + 1 << "\n";
    x0[it->first] ^= x0[it->second];
  }
  return 0;
}